From: "Dirk Van de moortel" <dirkvandemoortel@ThankS-NO-SperM.hotmail.com>
Subject: Re: Lion and Christian (Was:Hunter and deer)
Date: Wed, 31 Jan 2001 17:22:58 GMT
Newsgroups: sci.math
Summary: Pursuit problems (Besicovich's optimal strategy for the Christian)

"Virgil" <Vmhjr@frii.com> wrote in message
news:Vmhjr-8475CA.13092030012001@news.frii.com...
> In article <955rs7$fr3$1@nnrp1.deja.com>, Ron Larham
> <larhamr@my-deja.com> wrote:
>
> > > You didn't say that the lion is a dumb chaser, i.e. that he can
> > > anticipate where the christian is going next. I'll assume that the
> > > lion is less intelligent than that Neandertal guy.
>
> The ultimate solution is valid if both the lion and the man are Mensa
> members, or even 4-sigma members. No limit is imposed on the
> intelligence of the loser. The winner need only know his/its best
> strategy and be able to impliment it.
>
> It was discovered by A. S. Besicovitch, and first published in 1953, but
> not by Besicovitch.

Guess what.... I was curious about this Besicovitch and did a google search.
Abd behold, after a while I found this site:

http://www.amsta.leeds.ac.uk/pure/staff/partington/personal/quintics.html

An extract:
| A few more curious games to conclude with, that have no connexion
| with the above.

| First, The Lion and the Christian. The lion, L, and the Christian,
| C, are in a circular arena (unit radius), and L is chasing C with
| the object of eating him. Both move at unit speed, and are
| conveniently regarded as points. If L is at the centre, and C
| stays on the circumference, then it is not hard to show that L
| catches C in a time of pi/2, travelling on a pursuit curve.

| However C can do better (as Besicovitch showed). He moves towards
| L to start with, then makes a sequence of zig-zag steps at right
| angles to the direction the lion is from him at the start of each
| step: if the step lengths are chosen to be (di), where the sum of
| the (di) is infinite (so he is travelling for infinite time) and
| the sum of the (di2) is small and finite (so that he doesn't end
| up on the circumference again), C can evade L indefinitely.


So it IS possible :-)

Dirk Vdm